suppose that a force with magnitude of 52.8 N is exerted at an angle of 18 degrees with the horizontal. this force would be the same as having forces which are exerted at
To find the horizontal and vertical components of the given force, you can use trigonometric functions. Let's calculate:
Horizontal component: The horizontal component of a vector is given by the magnitude of the vector multiplied by the cosine of the angle it makes with the horizontal.
Horizontal component = 52.8 N * cos(18°)
Vertical component: The vertical component of a vector is given by the magnitude of the vector multiplied by the sine of the angle it makes with the horizontal.
Vertical component = 52.8 N * sin(18°)
Let's calculate these values:
Horizontal component = 52.8 N * cos(18°) = 49.94 N (rounded to two decimal places)
Vertical component = 52.8 N * sin(18°) = 15.41 N (rounded to two decimal places)
Therefore, the force can be considered as having two components:
1. A horizontal force of approximately 49.94 N
2. A vertical force of approximately 15.41 N
Is that the whole question?
I'm guessing to break that force up into x and y components
Fx = 52.8cos18
Fy = 52.8sin18